Question: Simplify the following expression and state the condition under which the simplification is valid. $q = \dfrac{-k^2 + 1}{k^3 + 8k^2 + 7k}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ q = \dfrac {-1(k^2 - 1)} {k(k^2 + 8k + 7)} $ $ q = -\dfrac{1}{k} \cdot \dfrac{k^2 - 1}{k^2 + 8k + 7} $ Next factor the numerator and denominator. $ q = - \dfrac{1}{k} \cdot \dfrac{(k + 1)(k - 1)}{(k + 1)(k + 7)}$ Assuming $k \neq -1$ , we can cancel the $k + 1$ $ q = - \dfrac{1}{k} \cdot \dfrac{k - 1}{k + 7}$ Therefore: $ q = \dfrac{ -k + 1 }{ k(k + 7)}$, $k \neq -1$